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\newcommand{\Sch}{i\hbar\frac{\partial\psi(\mathbf{x},t)}{\partial t}= \frac{-\hbar^2}{2m}\frac{\partial^2\psi(\mathbf{x},t)}{\partial \mathbf{x}^2}+V(\mathbf{x})\psi(\mathbf{x},t)}
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\title {Quantum Mechanics Lecture 3}

\maketitle
\section{Goal: Motivate The Fundamental Equation of QM: Schr\"odinger's Equation}
\[
|\psi(x_i,t)|^2\sim prob(x_i) \mbox{ where time evolution implies }|\psi(x_i,t)|^2\Delta x \sim prob(x_i,x_i+dx)
\]
\subsection{CLUES}
\begin{enumerate}
\item Photo-electric effect\\
$E_{particle}=h\nu_{wave}=\hbar\omega_{wave}$
\item Double Slit Experiment\\
$P_{particle}=\cfrac{h}{\lambda_{wave}}=\hbar k_{wave}$. Please recall that $\hbar=\cfrac{h}{2\pi}$ and $k=\cfrac{2\pi}{\lambda}$.
\item Equation Is Compatible with Wave Superposition\\
If Waves 1 and Waves 2 are solutions, then Wave 1 + Wave 2 is also a solution. In general, we can take linear combinations of these waves.
\item Non-Relativist Mechanics\\
$E=\frac{1}{2}mv^2=\cfrac{P^2}{2m}\iff \hbar\omega=\cfrac{(\hbar k)^2}{2m}$
\end{enumerate}
\section{Quantum Wave Theory}
\subsection{First a detour}
\paragraph{Working with Classical Mechanics, consider the string represented in Figure 1. With no restoring force acting on it, this string is flat. With a restoring force, this string is curved. Let the y-axis be a measure of displacement. From here, let's derive a wave equation.}
\begin{eqnarray*}
\vec{F}&=&m\vec{a}\\
\downarrow \hspace{1mm} & &\hspace{5mm}\searrow\\
\cfrac{\partial^2y}{\partial x^2}&= &\cfrac{1}{w^2}\cfrac{\partial^2 y}{\partial t^2}
\end{eqnarray*}
Where w is the speed of the wave, this implies that $w^2=\cfrac{\partial x^2}{\partial^2y}\cfrac{\partial^2 y}{\partial t^2}$, and $w=\cfrac{\omega}{k}=\pm\nu\lambda$.
\subsection{And Back To The Topic At Hand}
\paragraph{We don't have a starting equation at the moment. Rather, let's start with the classical story. Suppose we don't know that $\vec{F}=m\vec{a}$, but we do know that by studying the wave of a string's motion, we can get $y(x,t)=\sin(kx-w t)$. What equation is this a solution to?}
\begin{itemize}
\item
\[
\cfrac{\partial y}{\partial x} = k\cos(kx -w t)
\]
\item
\[
\cfrac{\partial y}{\partial t}=-w\cos(kx-w t)
\]
\item Then we should notice that
\[
\cfrac{\partial y}{\partial x}= -\cfrac{k}{w}\cfrac{\partial y}{\partial t}
\]
But there is a defect! We are only looking at waves in one direction! We need to consider both directions of the wave.
\item
\[
\cfrac{\partial^2 y}{\partial x^2}=-k^2\sin(kx-wt)
\]
\item
Now this will look at waves in both directions.
\[
\cfrac{\partial^2 y}{\partial x^2}=\cfrac{1}{w^2}\cfrac{\partial^2y}{\partial x^2}
\]
\item
As $w^2=(\lambda\nu)^2=(\cfrac{w}{k})^2\Rightarrow w=\pm \lambda\nu$
\end{itemize}
\paragraph{Now we should think of $\W=A\sin(kx-wt)$ where we can find momentum via $P=\hbar k$ and energy via $E=\hbar w$. More broadly, we should see from our fourth clue that:}
\[ E=\cfrac{P^2}{2m} \Rightarrow w=\cfrac{\hbar k^2}{2m}\] 
where 
\[\cfrac{\partial^2 y}{\partial x^2}=-Ak^2\sin(kx-wt)\]
give us our $k^2$ value, while 
\[
w=\cfrac{\partial \W}{\partial t} = -Aw\cos(kx-wt)
\]. 
If this is beginning to look familiar, it should.
\paragraph{Going back to our third clue that waves superpose, let's consider}
\[
\W=A\sin(kx-wt)+B\cos(kx-wt)
\]
\[
\cfrac{\partial^2\psi}{\partial x^2}=-Ak^2\sin(kx-wt)-Bk^2\cos(kx-wt)
\]
\[
\cfrac{\partial \psi}{\partial t}=Aw\cos(kx-wt)+Bw\sin(kx-wt)
\]
For $\cfrac{\partial^2\psi}{\partial x^2}$ to be equal (or proportional) to $\cfrac{\partial \psi}{\partial t}$, $A^2=-B^2$. If we choose $A^2=-B^2$, then we will relate the first order time derivative to the second order position derivative. We also learn that $A=\pm iB$. We should also note that the overall normalization of $\psi$ will always be chosen so that the computed probability makes sense.
\paragraph{We can conclude that if we have a particle}
\paragraph{.}\begin{tabular}{|ccc|}\hline Particle& $\mapsto$ & Wave\\\hline E & & w=$\cfrac{E}{A}$\\ P && $k=\cfrac{P}{\hbar}$ \\ \hline \end{tabular} $\Longrightarrow \W=\cos(kx-wt)-i\sin(kx-wt)=e^{i(kx-wt)}.$\\
\[
\cfrac{\partial^2\psi}{\partial x^2}=-ik^2\sin(kx-wt)-k^2\cos(kx-wt)
\]
\[
\cfrac{\partial \psi}{\partial t}=-w\cos(kx-wt)+w\sin(kx-wt)
\]
\[
\cfrac{\partial^2\psi}{\partial x^2}=\cfrac{-ik^2}{w}\cfrac{\partial \psi}{\partial t}=-i\cfrac{2m}{\hbar}\cfrac{\partial \psi}{\partial t} \Rightarrow \cfrac{-\hbar^2}{2m}\cfrac{\partial^2 \psi}{\partial x^2}=i\hbar\cfrac{\partial \psi}{\partial t}
\]
And now we should see that we have Schr\"odinger's equation in the case that the potential is zero.
\subsection{What do we now know?}
\begin{enumerate}
\item
We are forced into having wave functions that are complex
\item
This is nevertheless compatible with our probabilistic interpretation as $|\W|^2$is a real value
\item
For a particle with a given E and P, $\W=e^{i(kx-wt)}$. If we're given $\lambda$, then we know $P=\hbar k=\cfrac{h}{\lambda}$ If $|\W|^2=1$, then we find that our $\int\limits_{-\infty}^{\infty}|\W|^2=\infty$. In other words, complete certainty about momentum will give us complete uncertainty about the particle's location.
\item And all of this implies that thee is an equal probability that the particle will be found at ANY location. Since Schr\"odinger's equation is a linear differential equation, it has the property that if $\W_1$ and $\W_2$ both satisfy S.E., then $c_1\W_1+c_2\W_2$ will also satisfy S.E.
\end{enumerate}
\paragraph{Nowhere in this equation is there $\W^n$ as an acceptable solution. What matters is that as long as $\int\limits_{-\infty}^{\infty}|\W|^2\,dx \ne \infty$, $\W$ can be normalized and serve to be a description of the particle in question. What we need to do next is generalize the free particle Schr\"odinger equation (one with no potential V(x)) to a particle that is being acted on by a force.}
\subsection{CLUE 5}
If the particle encounters a \emph{potential}, such as gravity, EM, or whatever have you, recall that KE+PE-F; KE+V=E; and KE=(E-V(x)). All of this implies that $-\cfrac{\hbar^2}{2m}\cfrac{\partial^2\psi}{\partial x^2}=(E-V(X))\W$, where $E=i\hbar\cfrac{\partial \psi}{\partial t}$, as we've shown elsewhere. All of this yields
\[
\Sch
\]
\section{What We Know So Far}
\begin{itemize}
\item\textbf{Single Particle, Free Schr\"odinger Equation}
\[
i\hbar\cfrac{\partial \W}{\partial t}=-\cfrac{\hbar^2}{2m}\cfrac{\partial^2\W}{\partial x^2}
\]
\item\textbf{Single Particle Acted On By A  Force Described by V}
\[
\Sch
\]
\item\textbf{Single Particle in 3-D, Free Schr\"odinger Equation}
\[
\cfrac{i\hbar\partial\psi(x_1,x_2,x_3,t)}{\partial t}=-\cfrac{\hbar^2}{2m}\nabla^2\psi(x_1,x_2,x_3,t)
\]
\item\textbf{Single Particle in 3-D with Potential}
\[
\cfrac{i\hbar\partial\psi(x_1,x_2,x_3,t}{\partial t})=-\cfrac{\hbar^2}{2m}\nabla^2\psi(x_1,x_2,x_3,t)+V(x_1,x_2,x_3)\psi
\]
\end{itemize}
\end{document}